Differentiate ${\log _{10}}\left( {\sin x} \right)$ with respect to $x$.

Hint: To solve these types of questions where one function is inside the other function, we have to do the differentiation of these functions with the help of chain rule. Chain rule’s general form for finding the derivative of complex function like$f\left( x \right)$is
$\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = \dfrac{d}{{dx}}\left[ {f\left( u \right)} \right] \times \dfrac{{du}}{{dx}}$, where$u = g\left( x \right)$

Complete step-by-step solution:
Before starting to differentiate the given function, we should know about the chain rule in differentiation. In order to compute the derivative of a composite function, we use the chain rule. The standard form of a composite function is$f\left( {g\left( x \right)} \right)$. For differentiating such complex functions, we use chain rule.
$\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = \dfrac{d}{{dx}}\left[ {f\left( u \right)} \right] \times \dfrac{{du}}{{dx}}$, where$u = g\left( x \right)$
Since, the given function is also a composite function, we can use chain rule to differentiate it.
Let$f\left( x \right) = {\log _{10}}\left( {\sin x} \right)$. Now, we have to differentiate the given function with respect to $x$. Here, to solve this problem we will use the chain rule of differentiation.
We know the following log rule: ${\log _a}\left( b \right) = \dfrac{{\ln \left( b \right)}}{{\ln \left( a \right)}}$. Using the same, we get
$ = \dfrac{d}{{dx}}\left( {\dfrac{{\ln \left( {\sin \left( x \right)} \right)}}{{\ln \left( {10} \right)}}} \right)$
Now, we will take the constants out: ${\left( {a.f} \right)^\prime } = a.f'$
$ = \dfrac{1}{{\ln \left( {10} \right)}}\dfrac{d}{{dx}}\left( {\ln \left( {\sin \left( x \right)} \right)} \right)$
Next, we apply the chain rule
$ = \dfrac{1}{{\sin \left( x \right)}}\dfrac{d}{{dx}}\left( {\sin \left( x \right)} \right)$
We know that$\dfrac{d}{{dx}}\left( {\sin \left( x \right)} \right) = \cos \left( x \right)$. Using it in the above equation, we get
$ = \dfrac{1}{{\ln \left( {10} \right)}} \times \dfrac{1}{{\sin \left( x \right)}} \times \cos \left( x \right)$
After simplifying, we get
$ = \dfrac{{\cot \left( x \right)}}{{\ln \left( {10} \right)}}$
Hence, the derivate of ${\log _{10}}\left( {\sin x} \right)$ with respect to $x$ is $\dfrac{{\cot \left( x \right)}}{{\ln \left( {10} \right)}}$.


Note: The given question was an easy one. We cannot use chain rule in every function. The necessary condition for the chain rule to be applied in the differentiation of $f\left( {g\left( x \right)} \right)$ is that both $f\left( x \right)$ and $g\left( x \right)$ should be differentiable. In our case, both the functions were differentiable, that’s why we could use chain rule.